Solve for equal roots: x^2+k (2x+k-1)+2=0
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Given, equation, x2 + K (4x + K – 1) + 2 = 0
x2 + 4Kx + K(K – 1) + 2 = 0
For real and equal roots, D = 0
⇒ D = b2 – 4ac = 0
⇒ (4K)2 – 4 [1] [ K(K – 1) + 2 ] = 0
16K2 – 4 [ K2 – K + 2 ] = 0
16K2 – 4K2 + 4K – 8 = 0
12K2 + 4K – 8 = 0
3K2 + K – 2 = 0
3K2 + 3K – 2K – 2 = 0
⇒ 3K ( K +1 ) – 2 ( K + 1) = 0
⇒ ( K +1 ) ( 3K – 2) = 0

x2 + 4Kx + K(K – 1) + 2 = 0
For real and equal roots, D = 0
⇒ D = b2 – 4ac = 0
⇒ (4K)2 – 4 [1] [ K(K – 1) + 2 ] = 0
16K2 – 4 [ K2 – K + 2 ] = 0
16K2 – 4K2 + 4K – 8 = 0
12K2 + 4K – 8 = 0
3K2 + K – 2 = 0
3K2 + 3K – 2K – 2 = 0
⇒ 3K ( K +1 ) – 2 ( K + 1) = 0
⇒ ( K +1 ) ( 3K – 2) = 0

diksha13122002:
It's 2x in question not 4x
Value of k is 2
oh....sorry but u think you have now known how to solve....
but i think.....
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